Python的socket.connect奇怪的行爲

Question

我不能google一下，並不能明白爲什麼這個代碼工作它的工作原理

Python 3.5.2 (default, Sep 28 2016, 18:08:09) 
[GCC 4.2.1 Compatible Apple LLVM 8.0.0 (clang-800.0.38)] on darwin 
Type "help", "copyright", "credits" or "license" for more information. 
>>> import socket 
>>> s = socket.socket(socket.SOCK_DGRAM) 
>>> s.connect(('127.0.0.1', 33000)) 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
ConnectionRefusedError: [Errno 61] Connection refused 
>>> s.family 
<AddressFamily.AF_INET: 2> 
>>> 
>>> 
>>> s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM) 
>>> s.family 
<AddressFamily.AF_INET: 2> 
>>> s.connect(('127.0.0.1', 33000)) 
>>> 

編輯的方式：

Python 3.5.2 (default, Sep 28 2016, 18:08:09) 
[GCC 4.2.1 Compatible Apple LLVM 8.0.0 (clang-800.0.38)] on darwin 
Type "help", "copyright", "credits" or "license" for more information. 
>>> s2 = socket.socket(socket.SOCK_DGRAM) 
KeyboardInterrupt 
>>> import socket 
>>> s = socket.socket(socket.SOCK_DGRAM) 
>>> s.connect(('127.0.0.1', 33000)) 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
ConnectionRefusedError: [Errno 61] Connection refused 
>>> s.family 
<AddressFamily.AF_INET: 2> 
>>> s.type 
<SocketKind.SOCK_STREAM: 1> 
>>> 

貌似這個是TCP套接字=）我猜Python做了一些檢查並丟棄了不正確的值，但沒有提供任何有關該技巧的信息。

Answer 1

從documentation：

socket.socket([family[, type[, proto]]]) 

隨着socket.socket(socket.SOCK_DGRAM)使用SOCK_DGRAM家庭即使這是一個類型。這會導致奇怪的問題，如你看到的那個。